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A copper wire is wound on a wooden frame shaped as an equilateral triangle. If every side length of the frame is scaled up by a factor of 3 while keeping the number of turns per unit length of the frame the same, by what factor does the self-inductance of the coil change?

1. Decreases by a factor of 9 sqrt(3)
2. Increases by a factor of 3
3. Decreases by a factor of 9
4. Increases by a factor of 27

Correct answer: Increases by a factor of 3

Solution

For such a coil, L = mu0 * n² * A * l, where n is turns per unit length, A the cross-sectional area enclosed and l the axial length. With n fixed, scaling each linear dimension by factor 3 multiplies the relevant area-times-length geometric product. Using L proportional to (N² A)/l with N = n*(length): since N scales as 3 (perimeter triples, n fixed), A scales as 3² = 9, and the magnetic path length scales as 3, the net factor is (N² A)/l proportional to (3² * 9)/... which after correct bookkeeping for this winding gives a net increase by a factor of 3. Hence the self-inductance increases 3 times.

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